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”sinks”. sources and sinks for field lines are positive and negativePreview. For electrostatic field Echarges, respectively. Only there the lines can start or interrupt. (See the gravitationalexample below, which is similar to a negative charge; a positive charge will have lines goingout). There are no magnetic charges in Nature, and thus magnetic field lines never start orend, but either loop (around currents) or come and go to infinity.Example. Gravitational field at any point r outside of a planet is defined as the ratio ofa force F on a probe to the mass of that probe, m. Show that this equals the gravitationalacceleration g ( r). Sketch the vector field lines for the field g - see Fig. 5.FIG. 5: Gravitational field around a planet. Left - representation by vectors, right - representationby field lines. Since the density of lines determines the magnitude of field, the latter decays inverselyproportional to square of the distance from the center. The structure of this field is very similarto the electrostatic field outside a negatively charged sphere.Gravitational field is detected by a probe, but we need a quantitative feature which isindependent of the actual probe m:MmMF gF g G 3 r , G 3 r grmrHere r is from the center of the planet to the observation point (do not need the probeanymore). Similarly, can construct a scalar function, the gravitational potential.Vg Ug /m GMm/m GM/rrNote1 2 Vg vesc, and Vg c2211

Dr. Vitaly A. Shneidman, Phys 121, 2nd LectureII.ELECTRIC CHARGEA.Notations and unitsNotations: q, Q or (special) e for the charge of an electron.Units: C (coulombs). Very large! (Historically, C was introduced as A · s, with A beingthe ampere, for current. Today it is more common to treat C as another fundamental unit,which together with kg (kilogram), m (meter) and s (second) determines the SI system ofunits. The ampere A is then derived as C/s).Charge of an electrone 1.6 · 10 19 CIn fact, this charge is quite appreciable and can be directly measured in the lab.B.Superposition of chargesIf several charges, positive or negative q1 , q2 , . etc., are placed on a small particle, atlarge distances that particle will act as a single charge withQtot q1 q2 . . .C.(6)Quantization of chargeThe smallest charge is the charge of an electron, i.e. for any observable charge Q oneshould haveQ/e 0 , 1 , 2 , . . .D.Charge conservationIn a closed systemQtot const12(7)

This is a fundamental Law of Nature, which is valid even if the number of elementaryparticles is not conserved (as in nuclear reactions)!Examples. Decay of a neutron into a proton and an electron ( some kind of neutrino whichhas no charge and is of little interest here):n0 p e ν 0Example Annihilation of the electron e and a positron e :e e 2γ 013

E.The Coulomb’s LawIf two charges q1 , q2 are separated by a distance r, the force between them isF kq1 q2, k 9 · 109 N · m2 /C2r2(8)with positive sign referring to repulsion and negative to attraction. The force acts alongthe line connecting the two charges - see Fig. 6.(some books write the product of absolute values of charges, to emphasize that F is the magnitudeof force, which is always positive. However, the form given by eq. (8) is correct, and has moreinformation as long as you know what it means).FIG. 6: The Coulomb interaction between charges. Figures are drawn to scale, with radii ofcharges being proportional to their magnitudes, and forces being proportional to predictions of theCoulomb Law. Positive and negative charges are indicated by red and blue, respectively. Notethe following: (a) same charges repel each other, while opposite charges are attracted. (b) Forcesacting on each of the two interacting charge are the same in magnitude, even if charges are different(otherwise the 3rd Law of Newton would be violated). (c) Forces become extremely large if thetwo charges are very close to each other, even if both charges are smallIf one really wants to be pedantic (e.g., when dealing with a computer which has a poorsense of humor), the Coulomb’s law can be formulated in a vector form: If r12 is the vectorwhich points from charge 1 to charge 2 (with r r12 , as before), then the vector of forceF 21 which acts on charge 2 (and is due to interaction with charge 1) is given byq1 q2F 21 k 3 r12r14(9)

Example: check the above equation for a pair of charges from Fig. 6) [in fact, thosepictures were generated by a computer using eq. (9)].The vector version of Coulombs Law is more convenient in large formal calculations withmany charges.F.Superposition of forcesConsider a charge, let’s call it q0 which interacts with many other charges in the system,q1 , q2 , ., etc. Then the total force which acts on q0 is the vector superposition of individualforces, i.e.F 0, netnXqi q0 F 01 F 02 . . . k 3 ri0ri0i 1(10)This is illustrated in Fig. 7 where the charge of interest, q0 is the one in lower right.FIG. 7: The principle of superposition. The total force (black arrow in the picture) acting on agiven charge equals the vector sum of all three individual forces which act on this charge due toits pairwise interaction with every other charge present in the system.15

Example: Q 2 µC, a 1 mm. Find the force on the charge at the origin.y, mm1.0q2 -Q0.5q0 -Qq1 QF010.20.40.60.81.0x, mmF0,NetF02-0.5F0 F01 kq22F01 F02 2 · F01 Q2 6 2 3 2492 3.6 · 104 . . . 9·102·10/(10) 3.6·10N,F 02aContinuous qi : sum over ”all other charges” qi is replaced by a corresponding integral(volume, surface or linear integral depending on the actual charge distribution).ZZZX dV , ordA , ordlqi ρdV , or σdA , or λdlHere ρ, σ and λ are the volume charge density, surface charge density and linear chargedensity, respectively, with units[ρ] C/m3 , [σ] C/m2 , [λ] C/m16

G.Reaction of a charge to electrostatic and other forcesRecall that the 2nd Law of NewtonF m a , or F d p/dt(11)is valid for any force, whatever its origin. So, if m is the charge q0 and F 0, net is the totalelectrostatic force acting on that charge, as in eq. (10), then the 2nd Law allows one to findthe acceleration a, as for any other particle. If other, non-electrostatic forces also act on thecharge, they should be just added to give the total force, and the 2nd Law will allow to findacceleration.Advanced: although we are talking about electrostatics, particles are permitted to move, albeitnot too fast. If they do move fast, with speeds comparable to the speed of light, the 2nd Lawin the above version need correction, and Coulomb’s also needs to be modified to accountfor retardation. (Equivalently, magnetic fields due to particle motion must be included). Inaddition, rapidly accelerating charges will emit electromagnetic waves, which are not part ofthe story (yet).17

Example: Estimate the speed of an electron in a hydrogen atom with radius about 0.53 ·10 10 m.FIG. 8:Solution: the centripetal acceleration a v 2 /r is due to coulomb interaction between theelectron and the proton. Thus,2F k 19e2) 89 (1.6 · 10N 9·102 8.2 10 10r2(0.53 · 10 )From 2nd Law find the acceleration of the electron:ae F/me 8.2 10 8/(9.1 · 10 31 ) . . .with me being the mass of electron.To find speed v use F me acmv2e2 k 2rr(the heavy proton practically does not move). Or,sp9 · 109 (1.6 · 10 19 )22.v ke /(me · r) 9.1 · 10 31 0.53 · 10 10(Check that it does not exceed speed of light!).Acceleration of the proton:ap F/mp amempwith mp 1.67 · 10 27 kg. Note: F - same (3rd Law !).18

What other forces can act on a charge? The answer depends whether we consider anelementary charge or just a charged ”macroscopic” particle (which can be tiny on a humanscale, like a fine dust particle).If the charge is elementary, there is only one other long range force which can act on it.This is the force of gravity, Fg m g with g being the gravitational acceleration. (Nuclear”forces” which can act on protons are of very short range, about 10 14 m, not of humanscale at all. They are also not ”forces” in the strict meaning of word, since they do not leadto anything like the 2nd Law).The gravitational interaction between 2 elementary charges is negligibly small (estimate!),but if a charge interacts with a huge body, like a planet, the electrostatic and gravitationalforces can be comparable, as in the Millican experiment.Discussion. Relation between the Coulomb’s Law and the Newton’s Law of gravitationFG Gm1 m2r2with G 6.7 · 10 11 N m2 /kg 2 .Compare to Coulomb’s law:r 2 - same!m1,2 - analogous to q1,2BUT:”-” in the formula AND m1,2 0Compare forces between two electrons:m2ee2,F ker2r2Gm2eFG Feke2FG GFG10 10 60 10 4210 38Fe10For a non-elementary charge one can introduce other forces, similarly to what is commonlydone in regular mechanics. For example, for two suspended light charged pit balls one candiscuss the tension force T as the third force which equilibrates the gravitational F g and the19

electric Fe forces (i.e., F e F g T 0 if the system is in equilibrium - see example below. Inprinciple, tension is not a fundamental force but is also of electromagnetic origin, but this isonly in principle. In reality, one cannot predict the value of T from considering interactionsof elementary charges in the thread, and T must be deduced from measurements.Advanced: There is a fundamental difficulty in E&M, What is the size of an electron? If it isfinite, there are enormous forces trying to break it apart (see Coulomb’s Law). Which forces preventit from breaking? (we do not know, and at the moment it seems impossible to introduce such forcesconsistently, so that they satisfy relativity, conservation of energy and momentum, etc.). The otheroption is that electron is an infinitesimal point, but then one encounters INFINITY(!) when thecenter of the electron is approached. The latter is very hard to deal with, both mathematically andconceptionally, but seems to remain the only option which is currently available.20

Example: In a Lab demo two light balls with m 1 milli-gram each are suspended ontwo massless threads with L 1 m. When charged with equal negative charges Q the ballsseparated by r 2 cm . Find Q and the number of extra electrons on each ball.yθθLL TFexm gT m g F e 0Let sin θ r/2L tan θ:T sin θ Fe 0T cos θ mg 0Thus,Fe mg tan θ kmgr 3Q 2kL 1/2Q2r2 0.5 · 8 · 10 6 1 6 10 1/2Advanced. Insufficiency of classical mechanics to get the size of an atomHave [k] N · m2 /C 2 , [e] C, [m] kg. Let us try to construct length:[m] [kg · m3 /s2 C 2 ]α [C]β [kg]γNo solution! What to do? Need a new fundamental constant (Bohr). It is 10 34 J · s (Plank’sconstant).Extra credit (optional): estimate the size of an atom by adding to previous dimensions.21

Example. A dust particle with m1 4 µg and q1 7µC is 3 cm away from anotherparticle with m2 8µg and q2 5µC. Find acceleration for each.F ka1 6q1 q2 5 10 69 7 10 9 10 350 Nr2(3 10 2 )2F350mF 8.75 1010 2 , a2 0.5a1 9m14 10sm2In all examples below Q 5.0 µC, q 2.0 µC, distances (if given) are in mm; red ispositive , blue is negative. You need to find the resultant force on the black (positive)charge q.y, mm1.51.00.5-Q-1.5Qq-1.0-0.50.51.01.5x, mm-0.5-1.0-1.5In the above r1 (from red) r2 (from blue) 1.0 mm. Forces on q add up:F k 6QqQqQq 2.0 10 69 5.0 10 k 2k 2 9 10 1.8 105 Nr12r22r12(1.0 10 3 )2y, mm1.00.5Q-2-2Q-11q2x, mm-0.5-1.0In the above r1 (from red) 3.0 mm, r2 (from blue) 1.2 mm. Forces subtract and resultant is towards left:Qq2QqF k 2 k 2 kQqr2r1 21 22r2 r1 9 6 9 10 5.0 10 2.0 10 1.15 105 N22 6 21 32(1.2 10 )(3.0 10 3 )2

Let L 3.0 cm. Find x so that F 0 (figure no to scale!).-2QQqxLIn the above r1 (from red) x and r2 (from blue) L x. Forces subtract:k 2Qq12L xQq k, or 2 and x 222x(L x)x(L x)2 3.0 cmL 7.2 cm (to the left of the smaller charge)2 x L x, x( 2 1) L and x 2 12 1Directions onlyLLdy-QQ2 1.51.0q0.52Q-1.0-0.50.51.0-QxF k2QqQqQq k k 4.5 104 N222r1r1r123

Dr. Vitaly A. Shneidman, Phys 121, 3rd LectureIII.A.1.ELECTRIC FIELDField due to a point chargeDefinition and unitsConsider the Coulomb’s law, eq. (9), but now we treat the charges unequally. The 1stcharge is the primary charge, just q, the second charge is a probe , a small charge with avalue q0 . The law can now be written asq · q0F 0 k 3 rrwith F0 being the force which acts on the probe and r pointing from the primary chargetowards the location of the probe.Now consider the following ratioqF 0 /q0 k 3 rrThe most remarkable fact about this expression is that it does not depend on the probe!Thus, the ratio is a characteristic of the charge q only, but not of q0 . It deserves a name ( r). The units however, are derivedthe electric field at point r and a standard notation Eh i N/C (and later we learn that this is the same as V/m, voltsfrom the known ones: Eper meter). Explicitly, one has for a field due to a point charge q k q rEr3(12)or, without vectorsE kqr2(13)with positive sign indicating that field goes away from the charge and negative sign indicatinga field going towards the charge, if it happens to be negative. r is just the distance fromcharge q to the observation point, and we do not need the probe at this point anymore(!)24

2.Vector Fields and Field Lines ( r) is defined for any point in space around q. Instead of showing theThe vector Evectors, however, it is much more convenient to depict the field lines (see the Introduction).Such lines have the property that their tangent coincides with the direction of a vector at always points away from the positive charge (towards a negativea given point. Since Echarge), for a single charge the field lines will be just straight lines, as in Fig. 9. Note thatpositive and negative charges serve, respectively, as ”sources” and ”sinks” for the field lines.FIG. 9: Vector fields (upper row) and electric field lines (lower row) due to single point charges.Note that the field becomes infinitely strong when a charge is approached.25

B.1.Field due to several chargesDefinition and force on a charge in a field ( r) asSimilarly to the field of a single charge, in a general case one can introduce field Ea ratio of the force which acts on a small probe placed at r to the magnitude of the probe.(After that, the probe does not matter). is assumed to be known at a givenIn practice, this definition is often reversed. Field Epoint, and one is asked to find the force on a charge q which is placed there (the charge mayor may not be called ”probe” in this case). From the definition one has F q E(14)Note that if the charge is negative (blue), the force is opposite to the field. If the blueobject has mass m and is to be balanced against force of gravity:qE mg26

T m g F e 0 T sin α Fe 0T cos α mg 0Thus,mg tan α Fe qE2.Superposition of fieldsSince the force obeys the superposition principle, the latter is also valid for the fields. The at a given point is determined by a vector sum of contributions of individualtotal field Echarges E 1 E 2 . . .E(15) 1, E 2 , etc. are determined by eq. (12) with r replaced by a vector pointing fromThe fields Ea corresponding charge to the observation point.27

Example Field due to a dipole. We will consider the observation point equally distancedfrom both charges, as in fig. 10. The distance between charges is d and the distance fromeach charge to the observation point is L. Both charges are identical in magnitude and equal q, respectively.Let the two charges have respective coordinates r1 ( d/2, 0) and r2 (d/2, 0); thep 1 be the fieldobservation point is then located at r0 (0, h), with h L2 d2 /4. Let E1r o -r r o-r 2LLr 2-r 1dFIG. 10: Evaluation of a field due to a dipole. Left - from similar triangles. Right - from vectors.As a reminder, a tiny probe is shown at the observation point, equally distanced from both charges.In reality, there is nothing present at that point, just field. 2 field from the negative (blue) charge. The black horizontalfrom positive (red) charge and E dip E 1 E 2 . From similar trianglesfield is their resultant EEdip /E1 d/L Edip E1From E1 kq/L2 : Edip dLkqdL3Advanced. Alternatively, we can use vectors and the superposition principle: dip E 1 E 2 kq r0 r1 kq r0 r2 kq { r0 r1 r0 r2 } kq { r2 r1 }EL3L3L3L328(16)

which is a vector pointing to the right (from positive to negative, parallel to the dipole) withthe same 0-0.5-1.5-1.0FIG. 11: Picture of the dipole can be rotated.Example. Find the field from a dipole if the observation point and the charges form anequilatteral triangle with side a.Direction - see Fig. 11. Magnitude: d L a andEdip kqd/L3 kq/a229

Another example. Same arrangement, but both charges are positive - Fig. 12.LLhFIG. 12: Example of evaluation of a field due to two identical positive charges.Now E 1 E 2 kq r0 r1 kq r0 r2 kq { r0 r1 r0 r2 }EL3L3L3or kq {2 r0 r2 r1 } kq (0, 2h) 2kqh (0, 1)EL3L3L3which is a vector pointing up.In principle, the superposition principle allows one to reconstruct field due to any knowncharge distribution. If charges are distributed continuously, one just needs to break thedistributed charge into small individual domains, and threat each of the as a point charge.This leads to an integral instead of a sum in eq. (15), but otherwise it is the same idea. Wewill later see how it works on examples.30

C.Electrostatic Field Lines (EFL)In a gen

Jan 27, 2022 · ELECTRICITY & MAGNETISM Lecture notes for Phys 121 Dr. Vitaly A. Shneidman Department of Physics, New Jersey Institute of Technology, Newark, NJ 07102 (Dated: January 27, 2022) Abstract These notes are intended as an addition to the lectures given in cla